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Tetra Perp

Stage: 5 Challenge Level: Challenge Level:2 Challenge Level:2

tetrahedron
We can show that the edges $AD$ and $BC$ of a tetrahedron $ABCD$ are mutually perpendicular if and only if $AB^2 + CD^2 = AC^2 + BD^2$. A tetrahedron has three pairs of opposite edges. We have to prove that one pair are mutually perpendicular if and only if the sums of the squares of the lengths of the other two pairs are equal.

Edward of Graveney School, Tooting, London sent in an excellent solution to this question using vectors.
It is easiest to follow if the essential symmetry is made obvious by the notation. In order to do this we use the position vectors ${\bf a, b, c}$ and ${\bf d}$ for the vertices of the tetrahedron. The scalar product ${\bf a.a} = a^2$ gives the square of the length of the vector ${\bf a}$.

We know that

$$\eqalign { AB^2 + CD^2 &= ({\bf b - a})^2 + ({\bf d - c})^2 \cr &= a^2 + b^2 + c^2 + d^2 - 2 {\bf b.a} - 2{\bf d.c}}$$

and

$$\eqalign { AC^2 + BD^2 &= ({\bf c - a})^2 + ({\bf d - b})^2 \cr &= a^2 + b^2 + c^2 + d^2 - 2 {\bf c.a} - 2{\bf d.b}.}$$

Therefore

$$\eqalign { (AB^2 + CD^2)- (AC^2 + BD^2) &= 2({\bf c.a + d.b - b.a - d.c}) \cr &= 2({\bf c - b}).({\bf a - d}) }$$

The right hand side of this expression is just twice the scalar product of the vectors $\mathbf{BC}$ and $\mathbf{DA}$ and we know that the edges $BC$ and $DA$ are mutually perpendicular if and only if this scalar product is zero. The result follows.